leetcode_116번 - Populating Next Right Pointers in Each Node(트리, 폭 우선 검색)

Populating Next Right Pointers in Each Node

 
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Question
 Solution 
 
Given a binary tree

    struct TreeLinkNode {

      TreeLinkNode *left;

      TreeLinkNode *right;

      TreeLinkNode *next;

    }

 
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to  NULL .
Initially, all next pointers are set to  NULL .
Note:

You may only use constant extra space.

You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

 
For example,Given the following perfect binary tree,

         1

       /  \

      2    3

     / \  / \

    4  5  6  7

 
After calling your function, the tree should look like:

         1 -> NULL

       /  \

      2 -> 3 -> NULL

     / \  / \

    4->5->6->7 -> NULL

 
 
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이 문제는 내가 채택한 것은 광도 우선 검색이고, 대열로 하며, 동시에 매 줄의 결점 개수를 기록하는 것이다

#include<iostream>

#include<queue>

using namespace std;



//Definition for binary tree with next pointer.

struct TreeLinkNode {

		 int val;

		 TreeLinkNode *left, *right, *next;

		 TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}

		};



void connect(TreeLinkNode *root) {

	queue<TreeLinkNode*> temp_queue;	

	if(root==NULL)

		return;

	temp_queue.push(root);

	int row_size=1;

	while(!temp_queue.empty())

	{

		TreeLinkNode *p1=NULL;

		TreeLinkNode *p2=NULL;	

		while(row_size>0)

		{

			if(row_size==1)

			{

				p1=temp_queue.front();

				temp_queue.pop();

				p1->next=NULL;

			}

			else

			{

				p1=temp_queue.front();

				temp_queue.pop();

				p2=temp_queue.front();

				p1->next=p2;

			}

			row_size--;

			if(p1->left!=NULL)

				temp_queue.push(p1->left);

			if(p1->right!=NULL)

				temp_queue.push(p1->right);

		}

		row_size=temp_queue.size();

	}

}

int main()

{

	TreeLinkNode* root=(TreeLinkNode*)malloc(sizeof(TreeLinkNode));

	root->val=1;

	root->next=NULL;

	root->left=(TreeLinkNode*)malloc(sizeof(TreeLinkNode));

	root->right=(TreeLinkNode*)malloc(sizeof(TreeLinkNode));

	root->left->val=2;

	root->right->val=3;

	root->left->next=NULL;

	root->right->next=NULL;

	connect(root);

}